# Probability of a 4-digit number with all its digits different being a multiple of 11?

What is the probability of a 4-digit number of the form abcd(a,b,c,d are all different) is a multiple of 11?

(a) 0.1420
(b) 0.0480
(c) 0.0720
(d) 0.091


Can it be assumed that the ratio of 4-digit numbers of the form abcd which are multiples of 11 to that of all 4-digit numbers of the form abcd is 1? If yes, then is this right?

(((9999-1001)/11) + 1) / ( 9999 - 999 )= 91/1000

EDIT: This is what I intend to know

Is the proportion of the "all digits distinct" 4-digit numbers that are divisible by 11 among all 4-digit numbers divisible by 11 equal to the proportion "all digits distinct" 4-digit numbers among all numbers from 1001 to 9999?

Thanks to Andre Nicolas for stating this in his comment

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First question: obviously no. Second question: yes. –  Did Feb 20 '13 at 7:25
@Did Thank you. Please can you explain how it is solved? (That is not my solution) –  segmentation_fault Feb 20 '13 at 7:28
Sure I could. You first: what did you do? (You might be aware that to explain what one did is part of the recommended modus operandi on the site.) –  Did Feb 20 '13 at 7:29
Yes. There are 819 multiples of 11 between 1001 and 9999. And the total number of 4-digit numbers are 9000. But I'm stuck at how to proceed with the digits being different though this gives the stated answer. –  segmentation_fault Feb 20 '13 at 7:31
Would it help to know that $a+c$ and $b+d$ must be equal, or differ by 11? –  DJohnM Feb 20 '13 at 7:45